# Trig Identity

1. Oct 12, 2005

### brandon26

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.

2. Oct 12, 2005

### arildno

Well, how is cos^2A-sin^2A related to cos2A?

3. Oct 12, 2005

### TD

And then for the RHS, try changing the tan and the csc to the corresponding sine and/or cosine expressions.

4. Oct 12, 2005

### brandon26

5. Oct 12, 2005

### brandon26

I got as far as simplifying the equation to 2cosA / cos2A. What now?

6. Oct 12, 2005

### Tom Mattson

Staff Emeritus
So far so good. To finish you'll need the following trig identity:

$$\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}$$

You'll want to try to make the LHS of your identity look like that. To do that you'll need to divide the numerator and denominator of your expression by some other expression, and you need to figure out which one that is.

Hint: Look at the $1$ in the denominator above. What would you have to divide $\cos^2(A)$ by to get a $1$?

Last edited: Oct 12, 2005
7. Oct 12, 2005

### arildno

The simplest way is now to see:
$$\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)$$

I'll leave the last step to you..

Last edited: Oct 12, 2005
8. Oct 12, 2005

### TD

Once you have this, substitute tan(2a) by sin(2a)/cos(2a) and use the double angle formula on sin(2a). After that, realize that csc(a) = 1/sin(a) and you should be there

9. Oct 12, 2005

### brandon26

tan2A (2cosA/sin2A) = tan2A (2cosA/ (2sinAcosA)) = tan2AcosecA.

Safe boys!